Assume that \(X\cong PSL_{n}(q)\) . Then \(n\geqslant 3\)  by Remark REF , and \(X_{x}\)  lies in a large maximal geometric subgroup \(Y\)  of \(X\)  by Theorem REF . Then \(Y\)  is one of the groups listed in Proposition 4.7 of [1]}. It follows from Theorem 3.5 of [2]} and Proposition
5.2.15.(i) of [3]}, that \(Y\notin \mathcal {C}_{1}(X)\)  and \(Y\notin \mathcal {C}_{2}(X)\cup \mathcal {C}_{5}(X)\) , respectively, since \(\left( \Phi _{nf}^{\ast }(p),\left|Y\right|\right) >1\) . For the same reason \((X,Y)\)  is not equal to \((PSL_{4}(5),2^{4}.A_{6})\)  or to \((PSL_{3}(4),PSU_{3}(2))\) . Thus, one of
the following holds:

\(Y\)  is a \(\mathcal {C}_{3}\) -subgroup of type \(GL_{n/t}(q^{t})\) ,
where \(t=2\) , or \(t=3\)  and either \(q\in \left\lbrace  2,3\right\rbrace  \) , or \(q=5\)  and \(n \)  is odd.

\(Y\in \mathcal {C}_{8}(X)\) .

Assume that Case (i) holds. Then \(Y\cong SL_{n/t}(q^{t})\cdot (q^{t}-1)(q-1)^{-1}\cdot t\)  by [3]}, Proposition 4.3.6.(II). Also \(X_{x}=Y\)  by Lemma REF (1) and by [3]}, Proposition 4.3.6.(I), for \(n\geqslant 13\)  and by [6]} for \(n<13\) .
Assume that \(t=3\) . Note that \(\left|X\right|\leqslant \left|\mathrm {Out(}X\mathrm {)}\right|^{2}\cdot \left|X_{x}\right|\cdot \left|X_{x}\right|_{p^{\prime }}^{2}\)  by Lemma REF (1). Also, since \(2f\leqslant q\,\)  implies \(\left|\mathrm {Out(}X\mathrm {)}\right|\leqslant q(q-1)\) , we get \(\left|X\right|\leqslant q^{2}(q-1)^{2}\cdot \left|X_{x}\right|\cdot \left|X_{x}\right|_{p^{\prime }}^{2}\) . This in turn yields the inequality
(4.11) of [7]} (see part (3) in the proof of Proposition 4.2), which
leads to \(n/t=q=2\) , or to \(n=t=3\)  and \(q=2,3\)  or 5. The former is ruled
out, since \(X\)  is not isomorphic to \(PSL_{6}(2)\)  by Lemma REF , whereas \((n,q)=(3,2)\)  is ruled out in Proposition , since \(PSL_{3}(2)\cong PSL_{2}(7)\) . Finally, in the remaining
cases, \([X:X_{x}]=144\)  and 4000 for \(q=3\)  or 5 respectively. Thus, only
the former is admissible, since \([X:X_{x}]\)  is a square, and hence \(k=12\) .
Assume that \(t=2\) . Then \(n\)  is even and \(X_{x}\cong SL_{n/2}(q^{2})\cdot (q+1)\cdot 2\) . Then \(\left|X_{x}\right|\leqslant 2(q+1)\frac{q^{n^{2}/2}}{q^{4}-1} \leqslant 2q^{n^{2}/2-2}\)  and \(\left|X\right|\geqslant q^{n^{2}-2}\)  by [1]}, Corollary 4.3.(i), hence \(k^{2}=[X:X_{x}]>q^{n^{2}/2}/2\) . On the other hand, if \(n\geqslant 8\)  then \(k+1\)  divides \((q^{n}-1)(q^{n-2}-1)\)  by Lemma REF (3) and
by [7]}, Lemma 3.6. Therefore, \(q^{n^{2}/2}/2<k^{2}<q^{2n-2}\)  and
hence \(q^{(n-2)^{2}}<4\) , which has no solutions for \(n \geqslant 8\) .
If \(n=6\) , then
\(k^{2}=\frac{q^{9}\left( q-1\right)\left( q^{3}-1\right)\left( q^{5}-1\right) }{2(6,q-1)}\text{.} \) 
On the other hand, \(k+1\mid \left|X_{x}\right|\cdot \left|\mathrm {Out(}X\mathrm {)}\right|\)  by Lemma REF , and hence \(k+1\mid (q^{2}+1)(q^{3}+1)(q+1)^{2} \cdot 3f\) , with \(k+1\)  odd. So, we get
\(k+1\leqslant (q^{2}+1)(q^{3}+1)(q+1)q\text{.} \) 
However, there are no solutions in (REF ) by comparing it with (REF ).
If \(n=4\) , then
\(k^{2}=(q^{2}+q+1)\frac{q^{4}(q-1)^{2}}{2(4,q-1)}\) 
and hence \(q^{2}+q+1\)  is a square, but this contradicts [10]}, A7.1.
Assume that Case (ii) holds. Then, by [3]}, Propositions 4.8.3.(II),
48.4.(II) and 4.8.5.(II), one of the following holds:

\(Y\cong PSp_{n}(q)\cdot [\frac{(q-1,n/2)(q-1,2)}{(q-1,n)}]\)  with \(n\geqslant 4\) .

\(Y\cong PSU_{n}(q^{1/2})\cdot [\frac{(q^{1/2}+1,n)c}{(q-1,n)}]\) , with \(c=\frac{q-1}{(q^{1/2}+1,(q-1)/(q-1,n))}\) , \(n\)  odd and \(n\geqslant 3\) .

\(Y\cong PSO_{n}^{-}(q)\cdot 2\) , with \(q\)  odd, and \(n\geqslant 4\) .

If \(X_{x}<Y\) , then \(G_{x}\)  is a type 1 novelty with respect to \(Y\)  by Lemma REF (2). However, this is impossible by [3]}, Tables 3.5.H–I for \(n\geqslant 13\)  and by [6]} for \(n<13\) . Thus \(X_{x}=Y\)  in Cases (I)–(III).
Assume that case (I) holds. Then \(\left|X_{x}\right|\leqslant 4q^{n(n+1)/2}\)  and \(\left|X\right|> q^{n^{2}-2}\) , by [1]},
Corollary 4.3.(i) and (iii) respectively. Hence \(k^{2}>q^{(n^{2}-n-4)/2}\) . On the other hand, if \(n\geqslant 8\)  then \(k+1\)  divides
\((q^{n}-1)(q^{n-2}-1)\)  by Lemma REF (3) and by [7]}, Lemma 3.6.
So \(q^{(n^{2}-n-4)/2}/4<k^{2}<q^{2n-2}\)  and we again reach a contradiction.
Thus, \(3\leqslant n<8\)  and hence \(n=4\)  or 6 as \(n\)  is even. If \(n=6\) , then
\(k^{2}=\frac{q^{6}\left( q-1\right) ^{2}\left( q^{2}+q+1\right) \left(q^{4}+q^{3}+q^{2}+q+1\right) }{(q-1,3)}\) 
and hence \(q^{4}+q^{3}+q^{2}+q+1\)  is a square. So \(q=3\)  by [10]}, A8.1. However, it does not fulfill (REF ).
If \(n=4\) , then \(k^{2}=\frac{q^{2}\left( q^{3}-1\right) }{(q-1,2)}\) . On the other hand, \(k+1\)  divides \(2f \cdot q(q^{2}-1)\)  by Lemma REF (3), and hence \(k+1\)  divides \(2f \cdot (q+1)\)  as \(k\)  is coprime to \(q\frac{q-1}{(2,q-1)}\) . So, \(k<2f \cdot (q+1)\) , which, compared to \(k^{2}=\frac{q^{2}\left( q^{3}-1\right) }{(q-1,2)}\)  yields a contradiction.
Assume that Case (II) holds. Then \(\left|X_{x}\right|< \frac{(q^{1/2}+1,n)(q^{1/2}-1)}{(q^{1/2}+1,n)}q^{\frac{n^{2}-1}{2}}<2q^{n^{2}/2}\)  and \(\left|X\right|\geqslant q^{n^{2}-2}\) , by [1]}, Lemma 4.2.(ii) and Corollary
4.3.(i) respectively. Hence \(k^{2}>q^{n^{2}/2-2}/2\) . On the other hand, if
\(n\geqslant 5\)  then \(k+1\)  divides \(q^{n/2}+1\)  by Lemma REF (3) and by
[7]}, Lemma 3.6, as \(n\)  is odd. Thus \(q^{n^{2}/2-2}/2<q^{n/2}+1\) , which has
no solutions for \(n\geqslant 5\) . Therefore, \(n=3\)  and hence
\(k^{2}=q^{3/2}(q+q^{1/2}+1)\frac{(q^{1/2}-1,3)(q+1)}{(q-1,3)}. \) 
So \(q+q^{1/2}+1\)  is a square, but this contradicts [10]}, A7.1.
Finally, assume that Case (III) holds. Then
\(k^{2}=\frac{q^{n^{2}/4}(4,q^{n/2}+1)\left( q^{n/2}-1\right) }{2 \rho (q-1,n)}\prod _{j=1}^{n/2-1}(q^{2j+1}-1)\text{,} \) 
with \(\rho =1,2\) , and hence \(q^{n-1}-1\)  divides \(k^{2}\) . Also \(\Phi _{(n-1)f}^{\ast }(p)\mid k^{2}\) , where \(\Phi _{(n-1)f}^{\ast }(p)>1\) , by [10]}, P1.7, since \(n\geqslant 4\)  and \(n\)  is even. Thus \(\Phi _{(n-1)f}^{\ast }(p)^{1/2}\mid \left|G_{B}\right|\) , as \(k\mid \left|G_{B}\right|\) .
Note that \(\left( \Phi _{(n-1)f}^{\ast }(p),\left|X_{x}\right|\right) =1\) , as \(X_{x}\cong PSO_{n}^{-}(q)\cdot 2\) , with \(q\)  odd and \(n\geqslant 4\) . Also, \(\left( \Phi _{(n-1)f}^{\ast }(p),\left|\mathrm {Out(}X\mathrm {)}\right|\right) =1\) , as \(\Phi _{(n-1)f}^{\ast }(p)\equiv 1 \pmod {(n-1)f}\)  by [3]}, Proposition 5.2.15.(i). Thus, \(\left( \Phi _{(n-1)f}^{\ast }(p),\left|G_{x}\right|\right) =1\)  and \(\left(\Phi _{(n-1)f}^{\ast }(p),\left[ G_{B}:X_{B}\right] \right) =1\) , as both \(\left[ G_{x}:X_{x}\right] \)  and \(\left[ G_{B}:X_{B}\right] \)  divide \(\left|\mathrm {Out(}X\mathrm {)}\right|\) . Then \(\Phi _{(n-1)f}^{\ast }(p)^{1/2}\mid \left|X_{B}\right|\)  and \(\Phi _{(n-1)f}^{\ast }(p)^{1/2}\mid \left[ X:X_{B}\right] \) , as \(\left|G_{x}\right|=\lambda (k+1)\left|G_{x,B}\right|\)  and \(\left|G_{B}\right|=k\left|G_{x,B}\right|\) . Therefore, \(\Phi _{(n-1)f}^{\ast }(p)^{1/2}\)  divides both \(\left|X_{B}^{^}\right|\)  and \(\left[ SL_{n}(q):X_{B}^{^}\right] \) . Moreover,
\(q\)  divides \(\left|X_{B}^{^}\right|\)  by Corollary REF (i) for \(n\geqslant 4\) , since \(\left|X_{B}\right|_{p}=\left|X_{B}^{^}\right|_{p}\) . Then \(X_{B}^{^}\)  lies in a
parabolic subgroup of type \(P_{1}\)  of \(SL_{n}(q)\)  by Proposition REF (2). Hence \(X_{B}\)  fixes a point of \(PG_{n-1}(q)\) . Therefore, \(\frac{q^{n}-1}{q-1}\)  divides \(\left[ X:X_{B}\right] \)  and hence \(b\) . Thus \(\Phi _{nf}^{\ast }(p)\mid b\) . It follows from (REF ) that, \(\Phi _{nf}^{\ast }(p)\)  is coprime to \(k\) . Thus \(\Phi _{nf}^{\ast }(p)\mid k+1\) , since \(b=k(k+1)\lambda \)  and \(\lambda \mid k\) . Then \(X_{x}\in \mathcal {C}_{3}(X)\) 
by Lemma REF , as \(X_{x}\)  is maximal in \(X\) . However, this is
impossible, since \(X_{x}\in \mathcal {C}_{8}(X)\) .
